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import Mathlib

example (x : ℝ) (h : 0 ≤ x) (hn : ∀ n : ℕ, n > 0 → x ≤ 1 / n) : x = 0 := by
  by_contra
  have : 0 < 2 / x := (lt_div_iff₀ (by grind)).mpr (by grind)
  have := hn (Nat.ceil (2 / x)) (Nat.ceil_pos.mpr this)
  suffices x > 1 / (Nat.ceil (2 / x)) by linarith
  have := Nat.le_ceil (2 / x)
  have : 1 / x < Nat.ceil (2 / x) := by grind
  have := inv_strictAnti₀ (by grind) this
  field_simp at this
  exact this


example (x : ℝ) (h : 0 ≤ x) (hn : ∀ n : ℕ, x ≤ 1 / n) : x = 0 := by
  by_contra
  have : 0 < 2 / x := (lt_div_iff₀ (by grind)).mpr (by grind)
  have := hn (Nat.ceil (2 / x))
  suffices x > 1 / (Nat.ceil (2 / x)) by grind
  have := Nat.le_ceil (2 / x)
  have : 1 / x < Nat.ceil (2 / x) := by grind
  have := inv_strictAnti₀ (by grind) this
  field_simp at this
  exact this


example (x : ℝ) (h : 0 ≤ x) (hn : ∀ n, x ≤ 1 / n) : x = 0 := by
  linarith [hn 0]



lemma blah3 (x : ℝ) (h : 0 ≤ x) (hn : ∀ n, x ≤ 1 / n) : x = 0 := by
  have := hn 0
  simp at this
  exact le_antisymm this h

#print blah3


lemma blah4 (x : ℝ) (h : 0 ≤ x) (hn : ∀ n, x ≤ 1 / n) : x = 0 :=
  le_antisymm (Eq.mp (congrArg (LE.le x) (div_zero 1)) (hn 0)) h


-- example (x : ℝ) (h : 0 ≤ x) (hn : ∀ n : ℕ, x ≤ 1 / n) : x = 0 := by

--   have h₁ : x ≤ 1 := by
--     have h₂ := hn 1
--     norm_num at h₂ ⊢
--     <;> linarith
--   have h₂ : x = 0 := by
--     by_contra h₃
--     -- We will show that if x ≠ 0, then there exists a natural number n such that x > 1 / n, contradicting hn.
--     have h₄ : 0 < x := lt_of_le_of_ne h (Ne.symm h₃)
--     -- Since x > 0, we can find a natural number n such that x > 1 / n.
--     have h₅ : ∃ n : ℕ, x > 1 / n := by
--       -- Choose n = ⌈2 / x⌉₊.
--       use ⌈(2 : ℝ) / x⌉₊
--       have h₆ : (x : ℝ) > 0 := by exact_mod_cast h₄
--       have h₇ : (⌈(2 : ℝ) / x⌉₊ : ℝ) ≥ (2 : ℝ) / x := Nat.le_ceil _
--       have h₈ : (1 : ℝ) / ⌈(2 : ℝ) / x⌉₊ ≤ 1 / ((2 : ℝ) / x) := by
--         apply one_div_le_one_div_of_le
--         · positivity
--         · linarith
--       have h₉ : (1 : ℝ) / ((2 : ℝ) / x) = x / 2 := by
--         field_simp
--         <;> ring
--         <;> field_simp
--         <;> linarith
--       have h₁₀ : (1 : ℝ) / ⌈(2 : ℝ) / x⌉₊ ≤ x / 2 := by linarith
--       have h₁₁ : x > 1 / ⌈(2 : ℝ) / x⌉₊ := by
--         have h₁₂ : (x : ℝ) > 0 := by exact_mod_cast h₄
--         have h₁₃ : (⌈(2 : ℝ) / x⌉₊ : ℝ) ≥ (2 : ℝ) / x := Nat.le_ceil _
--         have h₁₄ : (1 : ℝ) / ⌈(2 : ℝ) / x⌉₊ ≤ x / 2 := by linarith
--         have h₁₅ : x > 1 / ⌈(2 : ℝ) / x⌉₊ := by
--           by_contra h₁₆
--           have h₁₇ : (x : ℝ) ≤ 1 / ⌈(2 : ℝ) / x⌉₊ := by linarith
--           have h₁₈ : (⌈(2 : ℝ) / x⌉₊ : ℝ) ≥ (2 : ℝ) / x := Nat.le_ceil _
--           have h₁₉ : (x : ℝ) * ⌈(2 : ℝ) / x⌉₊ ≤ 1 := by
--             calc
--               (x : ℝ) * ⌈(2 : ℝ) / x⌉₊ ≤ (1 / ⌈(2 : ℝ) / x⌉₊) * ⌈(2 : ℝ) / x⌉₊ := by gcongr <;> linarith
--               _ = 1 := by field_simp [Nat.cast_ne_zero.mpr (by positivity : ⌈(2 : ℝ) / x⌉₊ ≠ 0)] <;> ring
--           have h₂₀ : (x : ℝ) * ⌈(2 : ℝ) / x⌉₊ ≥ (x : ℝ) * ((2 : ℝ) / x) := by
--             gcongr <;> linarith
--           have h₂₁ : (x : ℝ) * ((2 : ℝ) / x) = 2 := by
--             field_simp [h₆.ne']
--             <;> ring
--             <;> field_simp [h₆.ne']
--             <;> linarith
--           linarith
--         exact_mod_cast h₁₅
--       exact h₁₁
--     -- Obtain the natural number n such that x > 1 / n.
--     obtain ⟨n, hn'⟩ := h₅
--     have h₆ := hn n
--     -- This leads to a contradiction because x ≤ 1 / n by assumption, but we have x > 1 / n.
--     linarith
--   exact h₂